Question

A2on is oxidized to ao3 by kmno4 in acidic medium. If 1.34 mmol of a2on requires 32.2 ml of 0.05 m acidified kmno4 solution for complete oxidation. Find the value of n.

Answer 1

Answer: Value of n is $2$.

Explanation:

$A_{2} O_{n}$ is oxidized to $AO_{3} ^{-}$ by $KMnO_{4}$ as:

                                     $A_{2} O_{n} + KMnO_{4}$   →   $AO_{3} ^{-} + Mn^{+2}$

Oxidation states:           $n$              $+7$               $+5$        $+2$

Now per mole of A, number of electrons supplied are $= 5 - n$

When $A_{2} O_{n}$ → $2AO_{3} ^{-}$ , number of electrons supplied = $2 (5-n)$

When $MnO_{4}^{-}$ → $Mn^{2+}$, number of electrons reacted = $+7 - (+2) = 5$

Now for this oxidation reaction to take place,

Number of equivalents of  $A_{2} O_{n}$ = Number of equivalents of $KMnO_{4}$

 milli moles * n-factor of $A_{2} O_{n}$ = milli moles * n-factor of $KMnO_{4}$

$1.34 * 2 (5-n) = 5 * 32.2 * 0.05 \\5 - n = 3 \\n= 2$

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Answer 2

Answer:

The value of n is 2.

Explanation:

AO3 is known by the  Molecule name. ALLOSAMIDIN.

KMnO  4  has a IUPAC name called Potassium manganate(VII).

Mn  +2   chemical name called magnesium.

Let then oxidation state of A will become as  n in A2On.

The reaction for the following is:

A2On+KMnO4→AO3−+Mn+2

  n               +7             +5          +2

The concept of 5−n factor is the number of electrons supplied or it can be reacted per mole of the substance.

For per mole of A, the number of electrons supplied will become,

=5−n.

As A2On→2AO3− total electrons supplied will become,

=2×(5−n).

For Mn, number of electrons reacted will become,

+7−(+2)=5.

Therefore, the value of electrons is

1.34×2×(5−n)

=5×32.2×0.05.

= (5−n)=3.

n=2

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